NIMCET 2014 — Mathematics PYQ

Solution

Let $S=\sin 20^{\circ }\sin 40^{\circ }\sin 80^{\circ }$.

Using the identity $2\sin A\sin B=\cos (A-B)-\cos (A+B)$:

$\Rightarrow S=\frac{1}{2}\sin 40^{\circ }(2\sin 80^{\circ }\sin 20^{\circ })$

$\Rightarrow S=\frac{1}{2}\sin 40^{\circ }[\cos (80^{\circ }-20^{\circ })-\cos (80^{\circ }+20^{\circ })]$

$\Rightarrow S=\frac{1}{2}\sin 40^{\circ }[\cos 60^{\circ }-\cos 100^{\circ }]$

Substituting $\cos 60^{\circ }=\frac{1}{2}$:

$\Rightarrow S=\frac{1}{4}\sin 40^{\circ }-\frac{1}{2}\sin 40^{\circ }\cos 100^{\circ }$

$\Rightarrow S=\frac{1}{4}\sin 40^{\circ }-\frac{1}{4}(2\sin 40^{\circ }\cos 100^{\circ })$

Using the identity $2\sin A\cos B=\sin (A+B)+\sin (A-B)$:

$\Rightarrow S=\frac{1}{4}\sin 40^{\circ }-\frac{1}{4}[\sin (100^{\circ }+40^{\circ })-\sin (100^{\circ }-40^{\circ })]$

$\Rightarrow S=\frac{1}{4}\sin 40^{\circ }-\frac{1}{4}[\sin 140^{\circ }-\sin 60^{\circ }]$

$\Rightarrow S=\frac{1}{4}\sin 40^{\circ }-\frac{1}{4}\sin 140^{\circ }+\frac{1}{4}\sin 60^{\circ }$

Since $\sin (180^{\circ }-x)=\sin x$, then $\sin 140^{\circ }=\sin 40^{\circ }$:

$\Rightarrow S=\frac{1}{4}(\sin 40^{\circ }-\sin 40^{\circ })+\frac{\sqrt{3}}{8}$

$\Rightarrow S=\frac{\sqrt{3}}{8}$

Correct Option: 3